Sort Array By Parity

Posted on | In |
  • Source : Link
  • Author : Han
  • Date : 2022-01-04

Problems

Given an integer array nums, move all the even integers at the beginning of the array followed by all the odd integers.

Return any array that satisfies this condition.

Example 1

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Input: nums = [3,1,2,4]
Output: [2,4,3,1]
Explanation: The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

Example 2

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Input: nums = [0]
Output: [0]

###Solution

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class Solution {
    public int[] sortArrayByParity(int[] nums) {
        final int length = nums.length;
        for(int i = 0; i < length; i++) {
            if((nums[i] % 2) == 1){
                for(int j = i + 1; j < length; j++){
                    if ((nums[j] % 2) == 0) {
                        final int temp = nums[j];
                        nums[j] = nums[i];
                        nums[i] = temp;
                    }
                }
            }
        }
        return nums;
    }
}

It’s better solution

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class Solution {
    public int[] sortArrayByParity(int[] nums) {
        final int two = 2;
        return Arrays.stream(nums).boxed().sorted((a, b) -> Integer.compare(a % two, b % two))
                .mapToInt(i -> i).toArray();
    }
}